Elijah and Tom had some coins in the ratio of 3 : 2. Tom and Lee had some coins in the ratio of 10 : 9. Elijah sold 70 coins to Lee and ended up with half the number of coins Tom had. How many coins did Lee have in the end?
| Elijah |
Tom |
Lee |
| 3x5 |
2x5 |
|
| |
10x1 |
9x1 |
| 15 u |
10 u |
9 u |
The number of coins that Tom had at first is repeated. Make the number of coins that Tom had at first the same. LCM of 2 and 10 is 10.
| |
Elijah |
Tom |
Lee |
| Before |
15 u |
10 u |
9 u |
| Change |
- 70 |
|
+ 70 |
| After |
5 u |
10 u |
9 u + 70 |
Number of coins that Elijah had in the end
= 10 u ÷ 2
= 5 u
Number of coins that Elijah sold to Lee
= 15 u - 5 u
= 10 u
10 u = 70
1 u = 70 ÷ 10 = 7
Number of coins that Lee had in the end
= 9 u + 70
= 9 x 7 + 70
= 63 + 70
= 133
Answer(s): 133
